If z is complex number such that |z| ≥ 2

Question 1:- If z is complex number such that |z| ≥ 2 . then the minimum value of |z + 1/2|.

(a) is equal to 5/2

(b) lies int interval (1, 2)

(c) is strictly greater than 5/2

(d) is strictly greater than 3/2 but less than 5/2 [2014 main]

Solution:- |z| = 2

let z = x + iy

|x + iy| = 2

⇒ x² + y² = 4 . . . (i)

|z| ≥ 2 is the region on or outside circle whose centre is(0, 0) and radius is 2

The minimum value of |z + 1/2| is distance of z, which lies on the circle |z| = 2 from (-1/2, 0).

∴ minimum value of |z + 1/2| = Distance between (-1/2, 0) and (-2, 0)

= √(-2 + 1/2)² + 0 = 3/2

Geometrically:- Min |z + 1/2| = AD

Hence, minimum value of |z + 1/2| lies in the interval (1, 2)

Question 2:- Let complex numbers α and $1/\overline{\alpha}$ lies on circles $(x - x_0)^2 + (y-y_0)^2 = r^2$ and $(x - x_0)^2 + (y-y_0)^2 = 4r^2$ , respectively.

If $z_0 = x_0 + iy_0$ satisfies the equation $2|z_0|^2 = r^2 + 2$ , then |α| is equal to

(2013 Adv.)

Solution:- See full solution

Question 3:- If w = α + iβ where β ≠ 0 and z ≠ 1 satisfies that condition that $\left(\dfrac{w - \overline{w}z}{1 - z}\right)$ is purely real, then the set of values of z is